Question

Find the integration of $\int {{e^{ax}} \cdot \sin \left( {bx + c} \right)} dx$

Answer 1

Here, we will solve the given integral using integration by parts. We will use the ILATE rule and the formula to integrate the integral. According to the integration by parts, we apply the ILATE rule when the product of two functions is given to us. The integral of the two functions is taken where the first term is the first function and the second term is the second function.

Formula Used:
We will use the following formulas:
1.$\int {a \cdot b = a\int b - } \int {\dfrac{{dy}}{{dx}}a \cdot \int b }$
2.$\dfrac{{dy}}{{dx}}\sin x = \left( {\cos x} \right)$
3.$\dfrac{{dy}}{{dx}}\cos x = \left( { - \sin x} \right)$

Complete step-by-step answer:
Given an integral equation, ${\rm{I = }}\int {{e^{ax}}.\sin \left( {bx + c} \right)} dx$.
Now we will apply the ILATE rule to integrate the given integral.
Now, according to the ILATE rule, a Trigonometric function comes before an exponential function, hence, we would interchange the positions of the given two functions.
$\Rightarrow I = \int {\sin \left( {bx + c} \right) \cdot {e^{ax}}} dx$…………………………$\left( 1 \right)$
Applying the formula of integration, $\int {a \cdot b = a\int b - } \int {\dfrac{{dy}}{{dx}}a \cdot \int b }$ in equation $\left( 1 \right)$, we get
$\Rightarrow I = \sin \left( {bx + c} \right) \cdot \int {{e^{ax}}dx} - \int {\left[ {\dfrac{{dy}}{{dx}}\left( {\sin \left( {bx + c} \right)} \right) \cdot \dfrac{{dy}}{{dx}}\int {{e^{ax}}dx} } \right]} dx$
Now, solving further i.e. differentiating and integrating wherever required, we get
$\Rightarrow I = \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} - \int {\left[ {b \cdot \cos \left( {bx + c} \right)\dfrac{{{e^{ax}}}}{a}} \right]} dx$………………………………$\left( 2 \right)$
Now let’s see how we solved the above integral and derivative parts respectively:
$\int {{e^{ax}}dx}$
This can be solved by using a substitution method.
Let $ax = y$
Differentiate both sides with respect to $x$
$\Rightarrow a = \dfrac{{dy}}{{dx}}$
$\Rightarrow dx = \dfrac{{dy}}{a}$
Hence, solving the integral by substituting these values, we get,
$\int {{e^{ax}}dx} = \int {{e^y}\dfrac{{dy}}{a}}$
This can be written as:
$\int {{e^{ax}}dx} = \dfrac{1}{a}\int {{e^y}dy}$
Also, $\int {{e^y}dy} = {e^y} = {e^{ax}}$
Hence, $\int {{e^{ax}}dx} = \dfrac{{{e^{ax}}}}{a}$
We know that the derivative of $\sin x$ is $\cos x$ .
Hence, solving the derivative, we get
$\dfrac{{dy}}{{dx}}\sin \left( {bx + c} \right) = \cos \left( {bx + c} \right) \cdot b$
Now, in equation $\left( 2 \right)$ , taking out constants from the integral part, we get,
$\Rightarrow I = \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} - \dfrac{b}{a}\int {\left[ {\cos \left( {bx + c} \right) \cdot {e^{ax}}} \right]} dx$
Now, again using integration by parts where the first trigonometric part is the first function and the second exponential part is the second function, we get,
$\Rightarrow I = \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} - \dfrac{b}{a}\left[ {\cos \left( {bx + c} \right) \cdot \int {{e^{ax}}dx} - \int {\left[ {\dfrac{{dy}}{{dx}}\cos \left( {bx + c} \right) \cdot \int {{e^{ax}}dx} } \right]} dx} \right]$
Using the formula $\int {a \cdot b = a\int b - } \int {\dfrac{{dy}}{{dx}}a \cdot \int b }$ and solving further, we get
$\Rightarrow I = \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} - \dfrac{b}{a}\left[ {\cos \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} - \int {\left[ { - b \cdot \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a}} \right]} dx} \right]$
Here, we applied the fact that:
$\dfrac{{dy}}{{dx}}\cos x = \left( { - \sin x} \right)$
Now, opening the brackets and solving further, we get,
$\Rightarrow I = \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} - \dfrac{b}{a}\cos \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a} + \dfrac{b}{a}\int { - b \cdot \sin \left( {bx + c} \right) \cdot \dfrac{{{e^{ax}}}}{a}dx}$
Taking out the constants from the integral part, we get,
$\Rightarrow I = \dfrac{1}{a}\sin \left( {bx + c} \right) \cdot {e^{ax}} - \dfrac{b}{{{a^2}}}\cos \left( {bx + c} \right) \cdot {e^{ax}} - \dfrac{{{b^2}}}{{{a^2}}}\int {\sin \left( {bx + c} \right) \cdot {e^{ax}}dx}$
Now, from the equation $\left( 1 \right)$ , we know that the integrating part is $I$.
$\Rightarrow I = \dfrac{1}{a}\sin \left( {bx + c} \right) \cdot {e^{ax}} - \dfrac{b}{{{a^2}}}\cos \left( {bx + c} \right) \cdot {e^{ax}} - \dfrac{{{b^2}}}{{{a^2}}}{\rm{I}}$
Now, shifting it to the L.H.S. and taking $I$ common , we get
$\Rightarrow I\left( {1 + \dfrac{{{b^2}}}{{{a^2}}}} \right) = \dfrac{1}{a}\sin \left( {bx + c} \right) \cdot {e^{ax}} - \dfrac{b}{{{a^2}}}\cos \left( {bx + c} \right) \cdot {e^{ax}}$
Now, solving further, by taking common , we get
$\Rightarrow I\left( {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}} \right) = {e^{ax}}\left( {\dfrac{{\sin \left( {bx + c} \right)}}{a} - \dfrac{{b\cos \left( {bx + c} \right)}}{{{a^2}}}} \right)$
Now, dividing LHS and RHS by $\dfrac{{{a^2}}}{{{a^2} + {b^2}}}$ we get,
$\Rightarrow I\left( {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}} \right)\left( {\dfrac{{{a^2}}}{{{a^2} + {b^2}}}} \right) = {e^{ax}}\left( {\dfrac{{\sin \left( {bx + c} \right)}}{a} \cdot \left( {\dfrac{{{a^2}}}{{{a^2} + {b^2}}}} \right) - \dfrac{{b\cos \left( {bx + c} \right)}}{{{a^2}}} \cdot \left( {\dfrac{{{a^2}}}{{{a^2} + {b^2}}}} \right)} \right)$
Taking the denominator ${a^2} + {b^2}$ common in RHS and eliminating the same variables, we get,
$\Rightarrow I = \left( {\dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}} \right)\left[ {a\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right]$
But,
$I = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)} dx$
$\therefore \int{{{e}^{ax}}\cdot \sin \left( bx+c \right)}dx=\left( \dfrac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}} \right)\left[ a\sin \left( bx+c \right)-b\cos \left( bx+c \right) \right]$
Hence, this is the required answer.

Note: For solving this question, we need to know the concept and different formulas of integration and differentiation. Integration is the summation of all the discrete data whereas differentiation is a process of reducing a function in parts.